Answer:
<em>The main sources of error in the collection of data are as follows : Due to direct personal interview. Due to indirect oral interviews. Information from correspondents may be misleading.</em>
Answer:
Exit velocity
m/s.
Explanation:
Given:
At inlet:
![P_1=100 bar,T_=600°C,V_1=35m/s](https://tex.z-dn.net/?f=P_1%3D100%20bar%2CT_%3D600%C2%B0C%2CV_1%3D35m%2Fs)
Properties of steam at 100 bar and 600°C
![h_1=3624.7\frac{KJ}{Kg}](https://tex.z-dn.net/?f=h_1%3D3624.7%5Cfrac%7BKJ%7D%7BKg%7D)
At exit:Lets take exit velocity ![V_2](https://tex.z-dn.net/?f=%20V_2)
We know that if we know only one property inside the dome then we will find the other property by using steam property table.
Given that dryness or quality of steam at the exit of nozzle is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.
Properties of saturated steam at 80 bar
![h_f= 1316.61\frac{KJ}{Kg} ,h_g= 2757.8\frac{KJ}{Kg}](https://tex.z-dn.net/?f=h_f%3D%201316.61%5Cfrac%7BKJ%7D%7BKg%7D%20%2Ch_g%3D%202757.8%5Cfrac%7BKJ%7D%7BKg%7D)
So the enthalpy of steam at the exit of turbine
![h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}](https://tex.z-dn.net/?f=h_2%3Dh_f%2Bx%28h_g-h_f%29%5Cfrac%7BKJ%7D%7BKg%7D)
![h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}](https://tex.z-dn.net/?f=h_2%3D1316.61%2B0.85%282757.8-1316.61%29%5Cfrac%7BKJ%7D%7BKg%7D)
![h_2=2541.62\frac{KJ}{Kg}](https://tex.z-dn.net/?f=h_2%3D2541.62%5Cfrac%7BKJ%7D%7BKg%7D)
Now from first law for open system
![h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w](https://tex.z-dn.net/?f=h_1%2B%5Cdfrac%7BV_1%5E2%7D%7B2%7D%2BQ%3Dh_2%2B%5Cdfrac%7BV_2%5E2%7D%7B2%7D%2Bw)
In the case of adiabatic nozzle Q=0,W=0
![3624.7+\dfrac{35^2}{2000}+0=2541.62+\dfrac{(V_2)^2}{2000}+0](https://tex.z-dn.net/?f=3624.7%2B%5Cdfrac%7B35%5E2%7D%7B2000%7D%2B0%3D2541.62%2B%5Cdfrac%7B%28V_2%29%5E2%7D%7B2000%7D%2B0)
m/s
So Exit velocity
m/s.
Answer:
a)
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b) 3.1 secs
Explanation:
a) Determine the normal times in TMUs for these motion elements
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b ) Determine the total time for this work element in seconds
first we have to determine the total TMU = ∑ TMU = 86.4 TMU
note ; 1 TMU = 0.036 seconds
hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds
Answer:
Explanation:
This will be possible when setting them up in summer with a certain quantity of sag, they have already know that the cables won't be able to sag any further because of the heat. During winter, when the cables contract because of the cold weather, the sag will therefore be reduced, but much tension will not be put on the cables.